Wileys problems in organic chemistry 1

P1.9 Identify the number of lone pairs of electrons that are present in the following structures.. P1.19 In the following compounds, identify the number of hydrogen atoms present around

ISBN 81-265-5582-3

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Organic Chemistry is understood by reading the textbook, listening to lectures and memorizing name reactions with

reagents But perhaps most importantly, it is learnt by doing, that is, solving problems It is not uncommon for students

who have performed below expectation in JEE to explain that they honestly thought they understood the text and lectures

The diffi culty, however, lies in applying, generalizing, and extending the specifi c reactions and mechanisms they have

“memorized” to the solution of a very broad array of related problems In doing so, students will begin to “internalize”

Organic Chemistry to develop an intuitive feel for, and appreciation of, the underlying logic of the subject Acquiring that

level of skill requires but goes far beyond rote learning It is the ultimate process by which one learns to manipulate the

myriad of reactions and, in time, gains a predictive power that will facilitate solving new problems

Mastering Organic Chemistry is challenging It demands memorization, but then requires application of those facts to

solve real problems It features a highly logical structural hierarchy and builds upon a cumulative learning process The

requisite investment in time and effort, however, can lead to the development of a sense of self-confi dence in the subject,

an intellectually satisfying experience, indeed

Many excellent textbooks are available to explain the theory of Organic Chemistry; all provide extensive exercises

Better performing students, however, consistently ask for additional exercises It is the purpose of this book, then, to

provide supplementary problems and their solutions that reinforce and extend those textbook exercises

This book has reached its destination in fi ve years; three years used in collection of problems chapter-wise and two

years for correction/eliminating errors I have designed this book in such a manner that it will be useful for JEE aspiring

students If you go through previous years’ JEE problems, you will fi nd that objective type questions have subjective

nature, that is why it was decided to write an objective pattern book in a subjective way The beauty of this book is in the

solutions, which are at par with international level treatment Each and every question has detailed solution with reaction

sequence, bond cleavage and formation of products The book has separate chapters for Substitution and Elimination

reaction and Carbonyl alpha-Substitution and Carbonyl Condensation reactions, because JEE has always been framing

problems from these chapters

Arrangement is according to classical functional group organization, with each group typically divided into Reactions,

Conversions and Mechanisms To emphasize the vertical integration of the subject, problems in later chapters heavily

draw upon and integrate reactions learned in earlier chapters

It is desirable, but impossible, to write a problem book that is completely text-independent Most problems will follow

a similar developmental sequence, progressing from alkane/alkene/alkyne to aromatic to aldehyde/ketone to carboxylic

acid to enol/enolate to amine chemistry But within the earlier domains, placement of the basics, stereochemistry, SN/E

mechanisms, and other functional groups varies considerably The sequence is important because it establishes the

concepts and reactions that can be utilized in subsequent problems It is the intent of this problem book to follow a

consensus sequence that complements a broad array of Organic Chemistry textbooks Consequently, instructors utilizing

a specifi c textbook may on some occasion offer their students guidance on the corresponding problem book chapter and

select problems for practice

I acknowledge the blessings and support of my mother Smt Tileshwari Devi, father Shri Triveni Singh, wife Mrs

Sarika Singh and sons, Aayushman and Shauryaman

I would like to thank Dr Sarika Mehta Jain for her unconditional support to make this an error-free book She spent

a lot of time on this book, and solved each and every problem to ensure authentic and error free solutions I can say

without any hesitation that she is the mother of this book, without her support this book would have never reached to my

students, readers and teachers I also sincerely thank all the members of Wiley India team and especially Paras Bansal,

Anjali Chadha and Seema Sajwan in bringing out this book in such a nice form It is all made possible by the grace of

God Almighty I devote this book to the feet of God

At the end, constructive criticism and valuable suggestions from the readers are most welcome to make the book more

useful

Dr K Singhksingh.visiondirector@gmail.com

ABOUT THE AUTHOR

Dr K Singh has done his post-graduation in Chemistry with special

paper in Organic Chemistry and his PhD in Water Pollution from Veer

Kunwar Singh University, Ara, Bihar He is an accomplished faculty in

Chemistry, imparting quality education to JEE aspirants across north

India He has a vast teaching experience of more than 19 years and has

taught more than 60,000 students He has churned out thousands of

successful students in JEE, many of whom attained top hundred ranks

and studied in institutes of their choice Many of these students are

doing exceedingly well in the fi eld of research and in corporate sector

He is one of the very few faculties from Bihar who had the privilege

of teaching at Kota in early days He then went on to establish the fi rst

Coaching Institute at Patna, Bihar, which is based on Kota teaching

pattern As a leading educationist of Patna, he very successfully stopped

the migration of the students from Bihar to Kota

Dr K Singh is a member of American Chemical Society and has

been conferred with many awards Some of which are: “Bharat Gaurav

Award” by World Economic Progress Society; Rashtriya Siksha Ratna Award” by I.E.D.R.A and “Bhojpuri Academy

Samman” by Govt of Bihar He has also participated in the “National Conference on Excellence in Higher Education”

His name is counted among the most successful and respected faculty in the world of JEE coaching

CONTENTS

8 ALCOHOLS AND ETHERS 205

12 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 371

13 CARBONYL SUBSTITUTION AND CONDENSATION REACTIONS 423

15 BIOMOLECULES AND POLYMERS 475

THE BASICS 1

1.1 Hybridization, Molecular Formula

and Physical Properties

P1.1 State the formal charge over any atom that possesses it

in the following structures

(b) All the electron pairs in both the structures

(c) The orbitals that overlap to form the covalent bonds cated by arrows (1), (2), (3), (4)

indi-(d) The hybridization state of both oxygen atoms in Compound (A) and of double bonded nitrogen in Compound (B)?

P1.3 In the following pairs of compounds, identify

(a) Compound with higher boiling point and why?

(b) Compound with lower melting point and why?

P1.4 Carbene [ :CH2], a highly reactive species containing divalent carbon having no charge and surrounded by sextet of electrons It can exist in two forms:

PROBLEMS

Singlet carbene with two unshared electrons in the same

orbital and triplet carbene with unpaired electrons in different

orbitals and linear HCH bond angle

(a) Identify the orbital housing the unshared electrons in

sin-glet carbene and predict the HCH bond angle

(b) Identify the orbitals housing the two unshared electrons

in triplet carbene

P1.5 Would the carbon atoms indicated by arrows in each of

the following structures lie in the same plane?

(b) Identify the type of orbital (s, p, sp, sp2, sp3) in which the lone pairs of electrons on the atoms indicated by arrows (1), (2) and (3) are present

(c) In Compound (A), the bond between the carbonyl carbon and nitrogen lies between a single and a double bond, sug-gesting resonance structure Identify the type of orbital in which the lone pair of electrons on that nitrogen resides

(d) How many lone pairs of electrons are present in each of the given structures?

P1.7 Among the following pairs, identify the species with

higher molecular dipole moment

(a) CHCl3 or CFCl3 (b) CH3NH2 or CH3NO2(c) CO2 or SO2

P1.8 Identify the functional groups present in each of the

P1.9 Identify the number of lone pairs of electrons that are

present in the following structures

(b) Number of lone pairs of electrons present in each compound

P1.11 Compounds (A) and (B) have the following

struc-tures In both the compounds, identify

(A)

(B)(a) The molecular formulae

(b) The number of sp2 and sp3 carbon atoms present

(c) The number of lone pairs of electrons present

P1.12 The following compound shows presence of two nitrogen atoms In this structure, identify

(a) The nature of orbital in which the lone pair on each N atom reside

(b) The hybridization of each N atom in the compound

P1.13 In the following compounds:

(a) State the hybridization of the N atom?

(b) Which of these compounds is the most basic?

P1.14 In the structure of acetonitrile given below

(a) What is the hybridization of both the C atoms and N atom?

(b) In what type of orbital does the lone pair on N reside?

(c) Identify the type of orbitals used to form each bond

P1.15 In the structure of two naturally occurring compounds

given, identify the orbitals used to form the bonds indicated with the arrows

(b)

P1.16 In the structure of antibiotic given below, identify:

(a) The hybridization of the atoms marked with an arrow

(b) The number of π bonds in the antibiotic.

P1.17 In the structure of compound given below

(a) Considering all the bonds, which is the shortest C C

bond?

(b) Identify the shortest and longest C C single bond

(c) Explain why the bond lengths of the two C C single

bonds (1) and bond (2) are different

P1.18 In the following compounds, indicate the orbitals used

to form the bonds marked with arrows Also indicate the individual orbitals used to form the multiple bonds

P1.19 In the following compounds, identify the number of

hydrogen atoms present around the carbon atom cated with arrows

indi-(a)

(1) (2) (3)

(4)

(b)

1.2 Acids and Bases

P1.20 Label each of the following compounds as a Lewis

acid or a base

(a) (b) Cl3C+

(c) BCl3 (d) CH3 — Cl(e) (f) H3O+

P1.21 Arrange the following ions in order of increasing basicity

(a) (b)

P1.22 Complete the following reactions showing the

move-ment of electrons using curved arrows Identify the reactants as Lewis acids or Lewis bases and place the formal charge on atoms as appropriate

P1.23 Complete the following reactions showing the

move-ment of electrons using curved arrows Identify the reactants as Lewis acids or Lewis bases

Write the Lewis acid base reaction when NaH is added to

NH3 Comment on the strength of NaH as base

P1.25 For the structures of the compounds given below, write a Brønsted-Lowry equation for the reaction of the conjugate base of the compound with

(b) Write the reaction for the conjugate base of Compound (A) with acetaminophen (Compound B) and identify the weak and strong acids and bases

(B)

P1 27 Which compound among the following has the lowest

pKa value and why?

(a) C2H5OH (b) CH3COOH (c) H2O

P1.28 Arrange the following compounds in order of

increas-ing acidity(a) NH3, H2O, HF

P1.29 Identify the products in each of the following

acid-base reactions Label the acid and acid-base in the tants and the conjugate acid and base in the products formed

P1.30 In the following three compounds, account for the

observed values pKa of three different C H bonds

Write all possible structures of the conjugate base of the compounds

P1.32 Explain the following:

(a) The C–H bond of nitromethane, pKa = 1.0, is more acidic than most C–H bonds

(b) The C–H bond of 1,4-pentadiene (marked) is more acidic than the C–H bond of pentane

(c) The pKa values of isomers dimethyl ether (pKa = 40) and

ethanol (pKa = 16) are different

P1.33 In the following compounds, indicate the most acidic

proton and explain why

P1.34 Draw the conjugate acid forms of the compounds

given below and on the basis of their stability predict protonation of which (a) oxygen and (b) nitrogen atom

P1.35 Identify the number of resonance structures possible

for the following ions

(g)

(h)

P1.36 For each of the following ions, draw a resonance structure that is more stable than the one given

(a)

(b)

(c)

(d)

P1.37 On the basis of resonance explain the following:

(a) The pKa value of Compound (B) is lower than that of Compound (A)

(b) The pKa value of H1 in Compound (C) is only about 10

(C)(c) The molecular dipole moment of Compound (E) is larger than that for Compound (D)

(D) (E)

P1.38 Identify the type of orbital in which the electrons

specified by the arrows are present

P1.39 Identify the number of resonance structures possible

for the following

P1.40 Illustrate the movement of electrons in the

follow-ing reactions usfollow-ing curved and or half-headed curved arrows

(a)

(b) (c)

(d) (e)

(f)

1.1 Hybridization, Molecular Formula and Physical Properties

S1.1 Formal charge is the charge that each atom in the molecule would carry if the electrons in the bonds were divided equally

between the two atoms It considers all bonds as if they were non-polar or the difference in the electronegativity of the

atoms is not taken into consideration The formal charges in the given compounds can be represented as follows:

(b) The presence of electron pairs in the structures is indicated as follows

(c) The bonds indicated by the arrows are:

(1): C C bond in which both the carbon atoms are sp3 hybridized, so orbitals that overlap are sp3 sp3;

(2): C C bond in which one carbon is sp3 hybridized and the second carbon (of aromatic ring) is sp2 hybridized,so orbitals

that overlap are sp3 sp2; (3): C O bond in which both atoms are sp2 hybridized, so orbitals that overlap are: sp2 sp2

(4): C N bond in which carbon and nitrogen both are sp2 hybridized, so orbitals that overlap are: sp2 sp2

(d) Hybridization of both the oxygen atoms in Compound (A) is sp3 and of nitrogen indicated in Compound (B) is sp2

S1.3

(a) The amine with the following structure has higher boiling point because this isomer can form intermolecular hydrogen bond This

increase in the intermolecular attractive forces leads to an increase in its boiling point as compared to the other isomer

(b) In the two given dihydroxy alcohols, intramolecular hydrogen bonding is possible in catechol but not in hydroquinone Due

to intramolecular attractive forces, the relative melting point for catechol is lower

S1.4

(a) In singlet carbene, the two unshared electrons are present in sp2 orbital, so the bond angle is for HCH bond is about 120°

(b) A linear HCH bond angle implies that carbon is in sp hybridization therefore each unpaired electron occupies an

unhybrid-ized p orbital The spins of electrons are aligned as per Hund’s rule.

S1.5

(a) Not in the same plane (b) Not in the same plane (c) In the same plane (d) In the same plane

(e) In the same plane (f) Not in the same plane (g) In the same plane (h) In the same plane

(c) Since the bond between carbonyl carbon and nitrogen is weaker than the double bond and stronger than single bond, it

suggests the presence of resonance structure of the type shown below

This suggests that lone pair of electrons on nitrogen resides in p orbital to be able to resonate.

(d) The number of lone pairs in Compound (A) is 12 and in Compound (B) is 8

S1.7

(a) CHCl3 will have higher dipole moment than CFCl3 The bond moment due to C F bond is in opposite direction to

resultant dipole moment of three C Cl bonds, hence net dipole moment of molecule decreases whereas no such

cancellation effect is seen in CHCl3

(b) CH3NO2 will have higher dipole moment than CH3NH2 because there is permanent charge separation in the former

μ = c × d; so more the charge separation, more will be dipole moment.

(c) SO2 will have higher dipole moment than CO2 because the latter has linear structure in which the net dipole moment

becomes zero SO2 has bent structure so the dipole moment does not cancel out and is greater than zero

S1.8 The functional groups present are:

(a) Alkene, amide, amine, ester, ether

(b) Alkene, amine, arene, carboxylic acid, halide, ketone

(c) Alcohol, alkyne, arene

S1.9

(a) Number of lone pairs is 8

(b) Number of lone pairs is 6

(b) Compound (A) contains 8 sp2 and 6 sp3 hybridized carbons

Compound (B) contains 5 sp2 and 15 sp3 hybridized carbons

(c) In Compound (A), the number of lone pairs is 7

In Compound (B), the number of lone pairs is 10

S1.12

(a) In the given structure, nature of orbitals in which the lone pair reside are: sp2 orbital for N1 and sp3 orbitals for N2

(b) The hybridization of nitrogen atoms is: sp2 for N1 and sp3 for N2

S1.13

(a) In the given compounds, the hybridization of nitrogen atom is:

(b) In going from sp to sp2 to sp3 hybridization, the percentage of s character decreases and the bond, H A becomes less

acidic and A:− become more basic So the order of basicity is:

Increasing order of basicity

S1.14

(a) The structure of acetonitrile involves single (sigma) bond between C C and a triple (1 sigma and 2 pi) bonds between

C N So the first carbon atom is sp3 hybridized, second carbon atom is sp hybridized and nitrogen is sp hybridized All

C H bonds are σ bonds, in which hydrogen atoms use 1s orbitals.

(b) The lone pair on N resides in a sp hybridized orbital.

(c) The type of orbitals used to form the bonds are depicted as follows:

S1.15

(a) The orbitals involved in the bond (marked with arrows) formation in the compound are as follows:

For bond (1): Oxygen (sp3) – hydrogen (1s)

For bond (2): σ bond: Carbon (sp2) – oxygen (sp2)

π bond: Carbon (p) – oxygen (p)

For bond (3): Carbon (sp3) – carbon (sp2)

For bond (4): Carbon (sp3) – carbon (sp3)

For bond (5): Carbon (sp3) – oxygen (sp3)

(b) The orbitals involved in the bond (marked with arrows) formation in the compound are as follows:

For bond (1): Carbon (sp3) – oxygen (sp3)

For bond (2): Carbon (sp2) – carbon (sp3)

For bond (3): Carbon (sp3) – carbon (sp3)

For bond (4): Carbon (sp2) – hydrogen (1s)

S1.16

(a) In the structure of the antibiotic, the hybridization of the atoms marked with an arrow is as follows:

For (1): Carbon atom is sp2 hybridized; bonded to three other atoms (trigonal planar geometry)

For (2): Nitrogen atom is sp3 hybridized; bonded to three other atoms and the lone pair resides in the fourth orbital

( tetrahedral geometry)

For (3): Oxygen atom is sp2 hybridized; bonded with carbon atom and the lone pairs reside in the other two sp2 orbitals

For (4): Carbon atom is sp3 hybridized; bonded to four other atoms (tetrahedral geometry)

For (5): Carbon atom is sp2 hybridized; bonded to three other atoms (trigonal planar geometry)

(b) The number of π-bonds in the molecule are six as depicted in the structure below.

S1.17

(a) Among all the C C bonds, bond (3) is the shortest because it is a carbon-carbpn triple bond

(b) The C C bond (3) is the shortest and the strongest because of increased percent of s character [C(sp) C(sp)] The C C

bond (1) is the longest and the weakest

(c) The carbon-carbon single bond (1) is longer than style bond (2) as it is a C(sp3) C(sp3) bond whereas bond (2) is a

C(sp3) C(sp2) bond and is shorter due to increased s character.

S1.18 The orbitals used in the formation of the indicated bonds are:

(a)

For bond (1): Carbon (sp3) − hydrogen (1s)

For bond (2): Carbon (sp3) − Carbon (sp3)

(b)

For bond (1): Carbon (sp2) − hydrogen (1s)

For bond (2): Carbon (sp2) − carbon (sp2) (σ-bond)

Carbon (p) − carbon (p) (π-bond)

For bond (3): Carbon (sp2) − carbon (sp3)

(c)

For bond (1): Carbon (sp2) − carbon (sp3)

For bond (2): Carbon (sp2) − oxygen (sp2) (σ-bond)

Carbon (p) − oxygen (p) (π-bond)

(d)

For bond (1): Carbon (sp) − hydrogen(1s)

For bond (2): Carbon (sp) − carbon (sp) (σ-bond)

Carbon (p) − carbon (p) (π-bond)

Carbon (p) − carbon (p) (π-bond)

For bond (3): Carbon (sp) − carbon (sp2)

For bond (4): Carbon (sp3) − nitrogen (sp2)

S1.19 The bond line formula shows only the carbon atoms in the zig-zag form The intersection of two or more lines and ends

of the lines represent carbon atoms (unless some other atom is mentioned) The end of line denotes a methyl group and the junctions denote carbon atoms bonded to appropriate number of hydrogen atoms

(a) The given structure can be written as

For carbon (1): three hydrogen atoms

For carbon (2): two hydrogen atoms

For carbon (3): one hydrogen atom

For carbon (4): three hydrogen atoms

(b) The given structure can be represented as

For carbon (1): zero hydrogen atom

For carbon (2): one hydrogen atom

For carbon (3): two hydrogen atoms

For carbon (4): two hydrogen atoms

For carbon (5): one hydrogen atom

For carbon (6): three hydrogen atoms

1.2 Acids and Bases

S1.20 A Lewis acid is any ionic or molecular species that can accept a pair of electrons in the formation of a coordinate

cova-lent bond A Lewis base is any ionic or molecular species that can donate electron pair in the formation of coordinate covalent bond They generally contain a lone pair or a π-bond.

Lewis acid (contains electron defi cient H+ ion)

S1.21 To solve this problem, we will write the structures of the conjugate acids of the given bases and arrange them in

increas-ing order of acidity Since the strongest acid has the weakest conjugate base, the order of basicity of the given ions will

be the reverse of order of acidity

(a) The conjugate acids of the given ions arranged in order of increasing acidity:

Thus, the increasing order of basicity:

(d) The conjugate acids of the given ions arranged in order of increasing acidity:

Thus, the increasing order of basicity:

(e)

S1.24

(a) (CH3)2NH is a stronger base than CH3 O CH3 because nitrogen is more electron releasing than oxygen

(b) The strongest base that can exist in ammonia is amide anion H2N−, which is a conjugate base of ammonia The Lewis acid

base reaction is

S1.25

(a) The conjugate base of ammonia can completely remove proton (H1) of the compound and the reaction is represented as:

(b) The conjugate base of NH3 can completely remove the proton from the given compound The reaction can be quantitatively

(b) The reaction of conjugate base of ibuprofen with acetaminophen is as given below

S1.27 Acetic acid and phenol are more acidic among given compounds because the anion formed is stabilized by resonance

However, in case of CH3COO− anion, the negative charge is stabillized over two oxygen atoms So, CH3COOH has the

lowest pKavalue due to which it has weakest conjugate base

S1.28 This problem can be attempted first based on the trends in acidic and basic character across coloumns and rows and then

comparing the effect of resonance, hybridization and electron-withdrawing group to determine the relative strengths of the acids

(a) The acid strength of hydrides increases as we move from left to right in a row Thus, the increasing order of acidity is:

3 2

NH <H O HF<

(b) The acidic strength of hydrides increases down the group This is because the tendency of the central atom to donate its

lone pair decreases due to increase in size Thus, the basic character decreases and the acidic character increases So the

increasing order of acidity is:

HF < HCl < HBr(c) The increasing order of acidity is:

2 3

OH H O H O

− < < +

(d) Between N H and O H bonds, the acid strength increases across the row, so OH is more acidic Between O H and

S H bonds, the acid strength increases down the column, so SH is more acidic Thus, the increasing order of acidity is:

bonds, the acid strength increases across a row, so H Cl is more acidic Thus the increasing order of acidity is:

2 2

H O H S HCl< <

(g) The order of acidic strength can be determined by nature of substituent:

CH3CH2CH3: Weakly acidic due to presence of C H bond only

CH3CH2OH: Acidic due to presence of O H group

ClCH2CH2OH: Strongly acidic due to presence of O H group and electron withdrawing Cl group, that stabilizes the

negative charge on removal of proton

Thus the increasing order of acidic strength is:

3 2 3 3 2 2 2

CH CH CH <CH CH OH ClCH CH OH<

(h) The acidic strength can be compared on the basis of hybridization of carbon

CH3CH2CH2CH3: carbon atom sp3 hybridized

CH3CH CHCH3: carbon atom sp2 hybridized

HC CCH2CH3: carbon atom sp hybridized

On going from sp3 to sp2 to sp hybridization, the percentage of s character increases and the bond, H A becomes more

acidic So increasing order of acidity is:

(j)

(k)

S1.30 We draw the possible structures of the conjugate base of the given compounds by first identifying the most acidic site

and removing a proton

(a)

(b)

(c)

For (a): There is only one structure possible for the conjugate base in which resonance is not possible So it is least acidic

For (b): There are two resonance structures possible for the conjugate base and the negative charge is delocalized over two

carbon atoms

For (c): There are two resonance structures possible for the conjugate base and the negative charge is stabilized over one

carbon and one oxygen atom So it is the most acidic

S1.31 The structure of the conjugate acid of a compound can be drawn by first identifying the most basic site and then protonating it

(a) The pKa is low for the C−H bond in CH3NO2because there are three possible resonance structures for the conjugate base

and the negative charge is delocalized over two oxygen atoms

(b) The more acidic nature of the marked proton in 1,4-pentadiene as compared to the marked proton in pentane can be

explained on the basis of resonance structures:

For 1,4-pentadiene: The three possible resonance stabilized strcutures of the conjugate base increase the acidic strength.

For pentane: There is only one possible structure for the conjugate base, so less stable base and hence less acidic

(c) The different pKa values of isomers dimethyl ether and ethanol can be explained on the basis of their structures

In dimethyl ether, all six hydrogen atoms are present on carbon, that is, C H bonds In ethanol, fi ve hydrogen atoms are

present as C H bonds whereas one is present as O H bond Since O H bond is more acidic than C H bonds, the pKa

value of ethanol is lower and it is more acidic

S1.33

(a) The acidic protons in the given structure are likely to be protons present as O H and N H Their relative strength can be

determined on the basis of stability of the conjugate bases formed

Conjugate base on loss of proton (1): It is stabilized by two resonance structures and by delocalization of negative charge

over oxygen in one structure

Conjugate base on loss of proton (2): There is no resonance stabilization, so proton is less acidic than proton (1)

Conjugate base on loss of proton (3): There is no resonance stabilization but the negative charge is on oxygen